# Solving Leetcode Interviews in Seconds with AI: Palindrome Rearrangement Queries # Introduction In this blog post, we will explore how to solve the LeetCode problem "2983" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months. # Problem Statement > You are given a 0-indexed string s having an even length n. You are also given a 0-indexed 2D integer array, queries, where queries[i] = [ai, bi, ci, di]. For each query i, you are allowed to perform the following operations: Rearrange the characters within the substring s[ai:bi], where 0 <= ai <= bi < n / 2. Rearrange the characters within the substring s[ci:di], where n / 2 <= ci <= di < n. For each query, your task is to determine whether it is possible to make s a palindrome by performing the operations. Each query is answered independently of the others. Return a 0-indexed array answer, where answer[i] == true if it is possible to make s a palindrome by performing operations specified by the ith query, and false otherwise. A substring is a contiguous sequence of characters within a string. s[x:y] represents the substring consisting of characters from the index x to index y in s, both inclusive.   Example 1: Input: s = "abcabc", queries = [[1,1,3,5],[0,2,5,5]] Output: [true,true] Explanation: In this example, there are two queries: In the first query: - a0 = 1, b0 = 1, c0 = 3, d0 = 5. - So, you are allowed to rearrange s[1:1] => abcabc and s[3:5] => abcabc. - To make s a palindrome, s[3:5] can be rearranged to become => abccba. - Now, s is a palindrome. So, answer[0] = true. In the second query: - a1 = 0, b1 = 2, c1 = 5, d1 = 5. - So, you are allowed to rearrange s[0:2] => abcabc and s[5:5] => abcabc. - To make s a palindrome, s[0:2] can be rearranged to become => cbaabc. - Now, s is a palindrome. So, answer[1] = true. Example 2: Input: s = "abbcdecbba", queries = [[0,2,7,9]] Output: [false] Explanation: In this example, there is only one query. a0 = 0, b0 = 2, c0 = 7, d0 = 9. So, you are allowed to rearrange s[0:2] => abbcdecbba and s[7:9] => abbcdecbba. It is not possible to make s a palindrome by rearranging these substrings because s[3:6] is not a palindrome. So, answer[0] = false. Example 3: Input: s = "acbcab", queries = [[1,2,4,5]] Output: [true] Explanation: In this example, there is only one query. a0 = 1, b0 = 2, c0 = 4, d0 = 5. So, you are allowed to rearrange s[1:2] => acbcab and s[4:5] => acbcab. To make s a palindrome s[1:2] can be rearranged to become abccab. Then, s[4:5] can be rearranged to become abccba. Now, s is a palindrome. So, answer[0] = true.   Constraints: 2 <= n == s.length <= 105 1 <= queries.length <= 105 queries[i].length == 4 ai == queries[i][0], bi == queries[i][1] ci == queries[i][2], di == queries[i][3] 0 <= ai <= bi < n / 2 n / 2 <= ci <= di < n n is even. s consists of only lowercase English letters. # Explanation Here's a solution to the problem, along with explanations: * **High-Level Approach:** * **Character Counts:** For each query, count the character frequencies within the rearrangeable substrings `s[a:b+1]` and `s[c:d+1]`. * **Diff Calculation:** Calculate the difference in character counts between the first half (`s[0:n//2]`) and the reversed second half (`s[n//2:n]`). Also, update these counts based on the character counts in the rearrangement substrings. * **Palindrome Check:** If all character counts differences can be made zero by rearrangement, then the answer is `True`, otherwise `False`. * **Complexity:** * **Runtime:** O(q * n), where q is the number of queries and n is the length of the string. For each query, we iterate through relevant portions of the string. * **Storage:** O(1). We use a fixed-size array (26 elements) to store character counts. ```python def solve(): s = input() queries = eval(input()) n = len(s) half = n // 2 ans = [] for a, b, c, d in queries: diff = [0] * 26 # Calculate initial difference between first and reversed second half for i in range(half): diff[ord(s[i]) - ord('a')] += 1 diff[ord(s[n - 1 - i]) - ord('a')] -= 1 # Adjust based on rearrangements counts1 = [0] * 26 counts2 = [0] * 26 for i in range(a, b + 1): counts1[ord(s[i]) - ord('a')] += 1 for i in range(c, d + 1): counts2[ord(s[i]) - ord('a')] += 1 for i in range(26): diff[i] += (counts2[i] - counts1[i]) # Check if all differences are zero total_diff = 0 for val in diff: total_diff += abs(val) if total_diff == 0: ans.append(True) else: ans.append(False) print(ans) # Example Usage (for testing) # Input strings must be enclosed in quotes # Queries must be enclosed as a list of lists. Example: [[1,1,3,5],[0,2,5,5]] # s = "abcabc" # queries = [[1,1,3,5],[0,2,5,5]] # Expected Output: [True, True] # s = "abbcdecbba" # queries = [[0,2,7,9]] # Expected Output: [False] # s = "acbcab" # queries = [[1,2,4,5]] # Expected Output: [True] # s = "cbacba" # queries = [[0,2,3,5]] # Expected Output: [True] # s = "aabbcc" # queries = [[0,1,2,3]] # Expected Output: [True] # s = "aabbcc" # queries = [[0,1,4,5]] # Expected Output: [True] # s = "leetcode" # queries = [[0,2,4,6]] # Expected Output: [False] # Code ```python def solve(): s = input() queries = eval(input()) n = len(s) half = n // 2 ans = [] for a, b, c, d in queries: diff = [0] * 26 # Calculate initial difference between first and reversed second half for i in range(half): diff[ord(s[i]) - ord('a')] += 1 diff[ord(s[n - 1 - i]) - ord('a')] -= 1 # Adjust based on rearrangements counts1 = [0] * 26 counts2 = [0] * 26 for i in range(a, b + 1): counts1[ord(s[i]) - ord('a')] += 1 for i in range(c, d + 1): counts2[ord(s[i]) - ord('a')] += 1 for i in range(26): diff[i] += (counts2[i] - counts1[i]) # Check if all differences are zero total_diff = 0 for val in diff: total_diff += abs(val) if total_diff == 0: ans.append(True) else: ans.append(False) print(ans) ```